Enthalpy Changes and Standard Conditions — AQA A-level Chemistry

SPEC 3.1.4.1-3.1.4.2 • ENERGETICS

Enthalpy Changes & Calorimetry

Understanding energy changes in reactions and how to measure them experimentally

What is Energetics?

Chemistry is fundamentally about energy changes. When chemical bonds break and form, energy is either released to or absorbed from the surroundings.

Key Terms

System: The chemicals involved in the reaction

Surroundings: Everything else (the container, air, water bath, etc.)

Why This Matters

Understanding energy changes helps us: predict whether reactions will occur spontaneously, design efficient industrial processes, calculate fuel efficiency, and understand biological energy transfer.

Energy cannot be created or destroyed - it can only be transferred between the system and surroundings. This is the First Law of Thermodynamics.

Exothermic & Endothermic Reactions

Exothermic

Energy is released from the system to the surroundings.

Temperature of surroundings increases.

ΔH is NEGATIVE

Examples: combustion, neutralisation, respiration

Endothermic

Energy is absorbed by the system from the surroundings.

Temperature of surroundings decreases.

ΔH is POSITIVE

Examples: thermal decomposition, photosynthesis

Understanding the Diagrams

Exothermic reactions are thermodynamically possible because the enthalpy of reactants is higher than products. However, if the rate is too slow, the reaction may not occur - this is called being kinetically controlled.

Key Features of Enthalpy Profile Diagrams

The activation energy (Eₐ) is the minimum energy needed for a reaction to occur - it is measured from the reactants to the transition state (peak)

The transition state is the highest energy point on the reaction pathway

In exothermic reactions, products are lower in energy than reactants → ΔH is negative

In endothermic reactions, products are higher in energy than reactants → ΔH is positive

Substances are more stable when they have less enthalpy

01Worked Example

Classify each reaction as exothermic or endothermic and state the sign of ΔH:

(a) The neutralisation of hydrochloric acid with sodium hydroxide causes the temperature to rise from 22°C to 35°C

(b) Dissolving ammonium nitrate in water causes the temperature to drop from 25°C to 18°C

(c) The decomposition of calcium carbonate: CaCO₃(s) → CaO(s) + CO₂(g) has ΔH = +178 kJ mol⁻¹

a)

Temperature of surroundings increases → energy released to surroundings → Exothermic, ΔH is negative

b)

Temperature of surroundings decreases → energy absorbed by system → Endothermic, ΔH is positive

c)

ΔH is already given as positive (+178) → Endothermic (this is thermal decomposition)

Enthalpy Change (ΔH)

Important Concept

Actual enthalpy cannot be directly measured, only the change. This is why we always talk about enthalpy changes (ΔH) rather than absolute enthalpy values.

Definition

Enthalpy change (ΔH) is the heat energy change measured under conditions of constant pressure.

Unit: kJ mol⁻¹ (kilojoules per mole)

The symbol "Δ" (delta) means "change in". So ΔH = H(products) − H(reactants).

Standard Conditions

Standard enthalpy changes are measured under standard conditions:

Pressure: 100 kPa (approximately 1 atmosphere)

Temperature: A stated temperature, usually 298 K (25°C)

Concentration: 1 mol dm⁻³ for solutions

Physical states: Standard states (most stable form at 298 K and 100 kPa)

Notation

The superscript plimsoll symbol (⦵) is used after ΔH to denote standard conditions. For example, ΔH° or ΔH⦵ refers to the standard enthalpy change.

State Symbols Matter

It is important to specify the physical states of each species in an equation when dealing with enthalpy changes, as changes in state can cause very large changes of enthalpy. For example:

NaCl(s) → Na⁺(aq) + Cl⁻(aq) ΔH = +4 kJ mol⁻¹

NaCl(g) → Na⁺(g) + Cl⁻(g) ΔH = +500 kJ mol⁻¹

Types of Standard Enthalpy Change

You need to know the definitions for these four types of standard enthalpy change:

Standard Enthalpy of Reaction (ΔrH°)

Standard enthalpy change of reaction (ΔrH°) is the enthalpy change when a reaction occurs in the molar quantities shown in the chemical equation, under standard conditions.

Example: CaO(s) + H₂O(l) → Ca(OH)₂(s) ΔrH° = −65 kJ mol⁻¹

The Four Types

Standard enthalpy change of reaction (ΔrH°) - Enthalpy change when a reaction occurs in the molar quantities shown in the chemical equation, under standard conditions

Standard enthalpy change of formation (ΔfH°) - Enthalpy change when 1 mole of a compound forms from elements under standard conditions

Standard enthalpy change of combustion (ΔcH°) - Enthalpy change when 1 mole of a substance combusts fully in oxygen under standard conditions

Standard enthalpy change of neutralisation (ΔneutH°) - Enthalpy change when an acid and alkali neutralise under standard conditions, to form 1 mole of water

Why Use Standard Conditions?

Using standard conditions means these values will be the same regardless of who measures them. This allows scientists worldwide to compare and use the same data.

Standard Enthalpy of Combustion (ΔcH°)

Definition

Standard enthalpy of combustion (ΔcH°) is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions, with all reactants and products in their standard states.

Key Features

Always involves one mole of the substance being burned

Combustion is complete (forming CO₂ and H₂O, not CO)

ΔcH° values are always negative (combustion releases energy)

Products include CO₂(g) and H₂O(l) at standard conditions

How to Write Combustion Equations

Start with 1 mole of the substance being burned on the left

Add O₂(g) as the other reactant (leave coefficient until last)

Balance carbon: Count C atoms → same number of CO₂ molecules

Balance hydrogen: Count H atoms → half that number of H₂O molecules

Balance oxygen: Count total O atoms needed on right, then work out O₂ coefficient (can be fractional, e.g., 6½)

Add state symbols: Fuel's state at 298 K, O₂(g), CO₂(g), H₂O(l)

Example: Standard enthalpy of combustion of methane

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔcH° = −890 kJ mol⁻¹

Common Mistake

The coefficient of the substance being burned must be 1. If you see "2CH₄", this is NOT a standard enthalpy of combustion equation.

02Worked Example

Write the thermochemical equation for the standard enthalpy of combustion of:

(a) Ethanol, C₂H₅OH (ΔcH° = −1367 kJ mol⁻¹)

(b) Propane, C₃H₈ (ΔcH° = −2220 kJ mol⁻¹)

a)

Start with 1 mole of ethanol. Balance carbon → CO₂, then hydrogen → H₂O, then oxygen.

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) ΔcH° = −1367 kJ mol⁻¹

b)

Start with 1 mole of propane. C₃H₈ has 3 carbons → 3CO₂ and 8 hydrogens → 4H₂O.

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) ΔcH° = −2220 kJ mol⁻¹

Note: State symbols are essential! Ethanol is a liquid at 298 K; propane is a gas.

Standard Enthalpy of Formation (ΔfH°)

Definition

Standard enthalpy of formation (ΔfH°) is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.

Key Features

Always involves one mole of the compound being formed

Reactants are elements (not compounds)

Elements must be in their standard states: O₂(g), H₂(g), C(graphite), S(s), etc.

ΔfH° of any element in its standard state = 0 kJ mol⁻¹

ΔfH° can be positive OR negative

How to Write Formation Equations

Write 1 mole of the compound as the only product on the right

Identify the elements that make up the compound

Write each element in its standard state on the left:

Diatomic gases: H₂(g), O₂(g), N₂(g), F₂(g), Cl₂(g)

Carbon: C(graphite) - NOT diamond

Metals: Usually solid, e.g., Fe(s), Na(s)

Balance using fractions if needed (e.g., ½O₂, 1½H₂)

Add state symbol for the product (its state at 298 K)

Example: Standard enthalpy of formation of water

H₂(g) + ½O₂(g) → H₂O(l) ΔfH° = −286 kJ mol⁻¹

Fractional Coefficients

Unlike combustion equations, formation equations often require fractional coefficients (like ½O₂) because only one mole of product is formed.

Summary of Standard Enthalpy Changes

TypeSymbolAlways Exo/Endo?ReactionΔrH°BothFormationΔfH°BothCombustionΔcH°ExothermicNeutralisationΔneutH°Exothermic

03Worked Example

Write the thermochemical equation for the standard enthalpy of formation of:

(a) Ammonia, NH₃ (ΔfH° = −46 kJ mol⁻¹)

(b) Ethanol, C₂H₅OH (ΔfH° = −277 kJ mol⁻¹)

a)

Form 1 mole of NH₃ from elements N₂(g) and H₂(g) in standard states.

½N₂(g) + 1½H₂(g) → NH₃(g) ΔfH° = −46 kJ mol⁻¹

b)

Form 1 mole of C₂H₅OH from C(graphite), H₂(g), and O₂(g).

2C(graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) ΔfH° = −277 kJ mol⁻¹

Note: Carbon's standard state is graphite, NOT diamond. This matters!

Measuring Enthalpy Changes with Calorimetry

We can measure the energy transferred in a reaction as heat using an apparatus called a calorimeter. This allows us to determine enthalpy changes for reactions.

Specific Heat Capacity

Specific heat capacity (c) is the energy needed to increase the temperature of 1 g of a substance by 1°C (or 1 K).

For water: c = 4.18 J g⁻¹ K⁻¹

Two Types of Calorimetry

Combustion calorimetry - used to find the enthalpy change when a fuel is burned, i.e. the enthalpy change of combustion

Solution calorimetry - used to find enthalpy changes for reactions occurring in aqueous solutions, such as neutralisation, dissolution, or displacement

The Principle

Energy released (or absorbed) by a reaction is transferred to (or from) the surroundings, usually water. By measuring the temperature change of the water, we can calculate the energy transferred.

Combustion Calorimetry

Combustion Calorimetry Method

A weighed fuel sample is burnt underneath a metal canister (calorimeter) containing a known mass of water

As the fuel combusts, heat is transferred to the water. A thermometer measures the temperature change

The temperature change along with the mass and heat capacity of water are used to calculate the heat energy

Solution Calorimetry

Solution calorimetry can find enthalpy changes for reactions occurring in aqueous solutions, including:

Neutralisation - acid + alkali reactions

Dissolution - dissolving a solid in water

Displacement - e.g. zinc + copper sulfate solution

Solution Calorimetry Method

Add a measured volume of one reactant to a polystyrene cup calorimeter and record initial temperature

Add a measured volume of a second reactant and quickly seal the container with a lid

Carefully stir the mixture with the thermometer and record temperatures at regular intervals over time as heat is released or absorbed

Why Polystyrene?

Polystyrene cups are used because polystyrene is a better insulator, lightweight, waterproof, and has a low specific heat capacity. These properties reduce heat loss to the surroundings and minimise the amount of heat absorbed by the calorimeter itself.

The Calorimetry Equation

To measure enthalpy changes in chemical reactions, you need to use two key equations sequentially: first to calculate the heat energy change, and then to use that value to calculate the enthalpy change per mole.

Formula

q = m c Δ T

SymbolQuantityUnit

q

Heat energy transferredJoules (J)

m

Mass of water or solutionGrams (g)

c

Specific heat capacity of water4.18 J g⁻¹ K⁻¹

Δ T

Temperature change (T_final − T_initial)K or °C

Kelvin vs Celsius

The official unit for ΔT is Kelvin (K). However, the Celsius scale gives the same numerical temperature change, so °C can also be used.

For example: A change from 20°C to 35°C is a ΔT of 15°C, which equals a change of 15 K.

Finding the Mass for Solutions

For aqueous solutions, if the density is 1.00 g cm⁻³, then:

mass (g) = volume (cm³)

If mixing two solutions, add the volumes together for the total mass.

04Worked Example

When 0.50 g of ethanol is burned completely, the heat released raises the temperature of 200 g of water from 20.0°C to 33.5°C. Calculate the heat energy released.

(Specific heat capacity of water = 4.18 J g⁻¹ K⁻¹)

1

Identify the values: m = 200 g, c = 4.18 J g⁻¹ K⁻¹, ΔT = 33.5 − 20.0 = 13.5°C

2

Apply q = mcΔT: q = 200 × 4.18 × 13.5

3

q = 11,286 J = 11.3 kJ (to 3 s.f.)

Answer: 11.3 kJ of heat energy was released

Calculating Molar Enthalpy Change

The heat energy (q) calculated from calorimetry is the total energy transferred. To find the molar enthalpy change (ΔH), we need to scale to one mole of reactant.

Formula

Δ H = \frac{- q}{n}

Where n = number of moles of the limiting reactant

Important: The Negative Sign

The formula includes a negative sign because:

When there is a rise in temperature, the value for ΔH becomes negative suggesting that the reaction is exothermic

When the temperature falls, the value for ΔH becomes positive suggesting that the reaction is endothermic

The negative sign ensures the correct sign convention is followed automatically.

Step-by-Step Method for Calorimetry Calculations

Calculate ΔT: ΔT = T_final − T_initial

Identify m: Mass of solution in grams (volume × density, or just volume if density = 1 g cm⁻³)

Calculate q: q = m × c × ΔT (answer in joules)

Calculate moles (n): n = mass ÷ Mᵣ OR n = concentration × volume (in dm³)

Calculate ΔH: ΔH = −q ÷ n

Convert units: Divide by 1000 to get kJ mol⁻¹

05Worked Example

50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rises from 22.0°C to 28.8°C. Calculate the enthalpy of neutralisation.

(Assume the density of the solution is 1.00 g cm⁻³ and c = 4.18 J g⁻¹ K⁻¹)

1

Total volume = 50.0 + 50.0 = 100.0 cm³. Mass = 100.0 g (since density = 1.00 g cm⁻³)

2

ΔT = 28.8 − 22.0 = 6.8°C

3

q = mcΔT = 100.0 × 4.18 × 6.8 = 2842.4 J

4

Moles of HCl = 0.0500 dm³ × 1.00 mol dm⁻³ = 0.0500 mol (same for NaOH)

5

ΔH = −q ÷ n = −2842.4 ÷ 0.0500 = −56,848 J mol⁻¹ = −56.8 kJ mol⁻¹

Answer: ΔH = −56.8 kJ mol⁻¹ (The accepted value is −57.1 kJ mol⁻¹; our value is slightly less negative due to heat loss)

06Worked Example

1.30 g of zinc powder is added to 50.0 cm³ of excess copper(II) sulfate solution. The temperature rises from 21.0°C to 42.5°C. Calculate the enthalpy change for the reaction.

(Mᵣ of Zn = 65.4; c = 4.18 J g⁻¹ K⁻¹; assume density of solution = 1.00 g cm⁻³)

1

Mass of solution = 50.0 g; ΔT = 42.5 − 21.0 = 21.5°C

2

q = mcΔT = 50.0 × 4.18 × 21.5 = 4493.5 J

3

Moles of Zn = 1.30 ÷ 65.4 = 0.01988 mol (Zn is limiting as CuSO₄ is in excess)

4

ΔH = −q ÷ n = −4493.5 ÷ 0.01988 = −226,030 J mol⁻¹ = −226 kJ mol⁻¹

Reaction: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s) ΔH = −226 kJ mol⁻¹

Creating an Accurate Temperature-Time Graph

During a calorimetry experiment, the temperature fluctuates throughout - increasing rapidly as heat is released, reaching a peak, then decreasing gradually again.

The Problem

If we just subtracted the final and initial values, we would underestimate the temperature change by missing the peak and not accounting for heat lost while measuring.

Finding the True Temperature Change

Plot the data - Begin by plotting the recorded temperature against time

Establish baseline - Draw a line of best fit for the initial temperature readings before the reaction commences. This is your baseline

Post-reaction line - After the peak reaction temperature is reached, draw another line of best fit for the cooling period

Extrapolation - Extend both lines to intersect at the reaction's start time

True temperature change - The difference between the two extrapolated lines of best fit at the start of the reaction gives the accurate temperature change

Sources of Error in Calorimetry

Experimental enthalpy values are usually less exothermic (less negative) than the accepted literature values. Understanding why helps you evaluate data and improve experiments.

Main Sources of Error

Source of ErrorEffect on ΔHHow to MinimiseHeat loss to surroundingsLess negative than true valueUse insulation (polystyrene cup), lidIncomplete combustionLess negative than true valueEnsure excess oxygen supplyHeat absorbed by calorimeterLess negative than true valueUse calorimeter with low heat capacityEvaporation of volatile substancesAffects accuracy unpredictablyUse a lid, work quickly

Improving Accuracy

Insulation: Polystyrene cup reduces heat transfer to surroundings

Lid: Prevents evaporation and convective heat loss

Stirring: Ensures even heat distribution throughout the solution

Extrapolation: Use a temperature-time graph to find the true ΔT

Repeat experiments: Take an average to reduce random errors

Key Assumptions

In calculations, we assume:

The solution has the same density as water (1.00 g cm⁻³)

The solution has the same specific heat capacity as water (4.18 J g⁻¹ K⁻¹)

All heat from the reaction is transferred to the solution (no heat loss)

Practice

Question 1: When 0.85 g of propan-1-ol (C₃H₇OH, Mᵣ = 60.0) is burned completely, the heat released raises the temperature of 250 g of water from 18.5°C to 40.3°C. Calculate: (a) the heat energy released, and (b) the enthalpy of combustion of propan-1-ol. (c = 4.18 J g⁻¹ K⁻¹)

a)

ΔT = 40.3 − 18.5 = 21.8°C

•

q = mcΔT = 250 × 4.18 × 21.8 = 22,781 J = 22.8 kJ

b)

Moles of propan-1-ol = 0.85 ÷ 60.0 = 0.01417 mol

•

ΔcH = −q ÷ n = −22,781 ÷ 0.01417 = −1,608,000 J mol⁻¹ = −1610 kJ mol⁻¹

Answer: (a) 22.8 kJ, (b) ΔcH° = −1610 kJ mol⁻¹

Practice

Question 2:

(a) The standard enthalpy of combustion of butane, C₄H₁₀ (ΔcH° = −2877 kJ mol⁻¹)

(b) The standard enthalpy of formation of carbon dioxide (ΔfH° = −394 kJ mol⁻¹)

a)

Combustion of butane (1 mole burns in oxygen):

C₄H₁₀(g) + 6½O₂(g) → 4CO₂(g) + 5H₂O(l) ΔcH° = −2877 kJ mol⁻¹

b)

Formation of CO₂ (1 mole formed from elements):

C(graphite) + O₂(g) → CO₂(g) ΔfH° = −394 kJ mol⁻¹

Practice

Question 3: 5.35 g of ammonium chloride (NH₄Cl, Mᵣ = 53.5) is dissolved in 100 cm³ of water. The temperature falls from 21.0°C to 15.2°C. Calculate the enthalpy of solution. (Assume density = 1.00 g cm⁻³, c = 4.18 J g⁻¹ K⁻¹)

1

Mass of solution ≈ 100 g (we assume the mass of water, ignoring the solid)

2

ΔT = 15.2 − 21.0 = −5.8°C (temperature falls, so ΔT is negative)

3

q = mcΔT = 100 × 4.18 × (−5.8) = −2424.4 J

4

Moles of NH₄Cl = 5.35 ÷ 53.5 = 0.100 mol

5

ΔH = −q ÷ n = −(−2424.4) ÷ 0.100 = +24,244 J mol⁻¹ = +24.2 kJ mol⁻¹

Answer: ΔH = +24.2 kJ mol⁻¹ (positive because dissolution of NH₄Cl is endothermic)

Summary

Key Definitions

TermDefinition

Enthalpy change (ΔH)Heat energy change at constant pressure

Standard conditions100 kPa, 298 K (usually), standard states

ΔcH°Energy change when 1 mol burns completely in O₂

ΔfH°Energy change when 1 mol forms from elements

Key Formulas

q = mcΔT ΔH = −q ÷ n

Exam Tips

Always include state symbols in thermochemical equations

ΔcH° is always negative; ΔfH° can be positive or negative

Remember the negative sign in ΔH = −q/n

Convert cm³ to dm³ (÷1000) when calculating moles from concentration

Give final answers in kJ mol⁻¹